Let \(\{a_n\}\) be a sequence of positive real numbers. Suppose that \(l = \lim_{n \to \infty} \dfrac{a_{n+1}}{a_n}.\) Which of the following is true?
Show Hint
- If \( l = 1 \), then \( \lim_{n \to \infty} a_n = 1 \)
- If \( l = 1 \), then \( \lim_{n \to \infty} a_n = 0 \)
- If \( l < 1 \), then \( \lim_{n \to \infty} a_n = 1 \)
- If \( l < 1 \), then \( \lim_{n \to \infty} a_n = 0 \)
The Correct Option is D
Solution and Explanation
Step 1: Recall the ratio test for sequences.
If \( \displaystyle \lim_{n \to \infty} \dfrac{a_{n+1}}{a_n} = l \), the behavior of the sequence depends on \( l \):
- If \( l < 1 \), terms \( a_n \) keep decreasing (approaching zero).
- If \( l = 1 \), the test is inconclusive (the limit could be 0, finite, or infinite).
- If \( l > 1 \), the terms increase indefinitely (diverge).
Step 2: Apply this understanding.
Given \( a_n > 0 \) and \( \dfrac{a_{n+1}}{a_n} \to l < 1 \), each term \( a_{n+1} = a_n \cdot l \) approximately becomes smaller. Hence, as \( n \to \infty \), \( a_n \to 0. \)
Step 3: Conclusion.
When \( l < 1 \), the sequence converges to \( 0 \). Therefore, the correct statement is: \[ \boxed{\lim_{n \to \infty} a_n = 0.} \]
Answer: Option (D)
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